报告主题

本报告为了解决以下几点问题：

  1. 分析社交媒体上对电影的评分与火爆程度（评论人数）和电影票房的关联。
  2. 分析在热门的社交媒体上哪些类型的电影更受欢迎，票房成绩更好。
  3. 分析在热门的社交媒体上哪些电影演员更受欢迎，上榜数量多。
  4. 基于支持向量机（SVM），建立电影票房评价的支持向量分类模型，并检验准确率。

数据来源

本报告对豆瓣【中国内地电影票房总排行】网站进行爬虫，主要采集以下几方面的数据，爬虫代码见代码部分的crawl.R。

分析思路

分析思路1

  • 电影评分Rating为十分制，存在一位小数，且1分、2分左右的电影数量少，因此可以将电影划分成低于3分、3-5分、5-6分、6-7分、7-8分、8-9分、高于9分这7个等级，统计得到每个等级对应的平均票房、社交媒体上的平均评论人数，绘制相关柱形图，分析平均票房与评分、平均评论人数与评分的相关关系。

分析思路2

  • 统计该排行中各类型Type上榜电影的数量，并统计各类型电影的总票房，绘制相关饼图，分析更受欢迎和更卖座的电影类型。因为提取到的电影类型Type是以诸如“喜剧/爱情/奇幻”等多类型存储，所以对应的该部电影在三个类别中均上榜，但在计算票房时需按照类别数将票房均分。

分析思路3

  • 统计该排行中各主演Actor的出现次数，并绘制词云，因出现1次的演员数量过多且意义不大，所以绘制出现次数2次及以上的主演的词云。

分析思路4

  • 将影响电影票房的因素作数字化与归一化处理，使用支持向量机（SVM）训练预处理后的电影相关数据，建立有关票房评价的分类模型，并检验分类模型的准确率。

计算过程

以下是数据分析和数据挖掘过程

  1. 基本统计分析
# Import packages 
library(recharts)
library(wordcloud2)
library(stringr)
library(plyr)
library(DT)
# Read the data
movies<-read.csv("..\\data\\movies.csv")
# Delete invalid information
movies<-na.omit(movies)
  1. 统计不同评分下电影的票房与社交媒体上的评价人数：
# ---------- Draw the histogram of Sales_avg and Appraise_avg in different ratings ----------
# Divide ratings into seven ranks 
rating_1<-subset(movies,Rating<=3)
rating_2<-subset(movies,Rating>3 & Rating<=5)
rating_3<-subset(movies,Rating>5 & Rating<=6)
rating_4<-subset(movies,Rating>6 & Rating<=7)
rating_5<-subset(movies,Rating>7 & Rating<=8)
rating_6<-subset(movies,Rating>8 & Rating<=9)
rating_7<-subset(movies,Rating>9)

# Build a new frame for different ratings about Sales_avg and Appraise_avg
rating_df<-data.frame(Rating=c("0~3","3~5","5~6","6~7","7~8","8~9","9~10"),
                      Sales_avg=c(mean(rating_1$Sales),mean(rating_2$Sales),
                                  mean(rating_3$Sales),mean(rating_4$Sales),
                                  mean(rating_5$Sales),mean(rating_6$Sales),
                                  mean(rating_7$Sales)),
                      Appraise_avg=c(mean(rating_1$Appraise),mean(rating_2$Appraise),
                                     mean(rating_3$Appraise),mean(rating_4$Appraise),
                                     mean(rating_5$Appraise),mean(rating_6$Appraise),
                                     mean(rating_7$Appraise)))
# Draw the histogram of Sales_avg in different ratings
Sales_eBar<-eBar(rating_df,~Rating,~Sales_avg)+
  eTitle(title="各评分电影平均票房数柱状图")+
  eLegend(show=FALSE)
Sales_eBar
# Draw the histogram of Appraise_avg in different ratings
Appraise_eBar<-eBar(rating_df,~Rating,~Appraise_avg)+
  eTitle(title="各评分电影平均评论数柱状图")+
  eLegend(show=FALSE)
Appraise_eBar
  1. 统计不同类型电影的上榜数量与所占总票房：
# ---------- Draw the pie chart of types and sales of types ----------
# Extract the type column and divide by /
type<-movies$Type
type<-as.character(type)
type<-strsplit(type,"/")
# Extract the sales column
sales<-movies$Sales

# Create two new lists of type and sales
type_list<-list()
sales_list<-list()
for(i in 1:nrow(movies)) {
  for(j in 1:length(type[[i]])) {
    # Take the every type of each movie and combine it into a new list
    add_list<-list(type[[i]][j])
    type_list<-c(type_list,add_list)
    # Divide the box office of each movies by number of types 
    add_list<-list(sales[[i]]/length(type[[i]]))
    sales_list<-c(sales_list,add_list)
  }
}

# Build two new frames for type and sales
type_df<-data.frame(unlist(type_list))
sales_df<-data.frame(unlist(sales_list))
# Insert the ID column into type_df and sales_df
ID<-c(1:nrow(type_df))
type_df<-data.frame(ID,type_df)
sales_df<-data.frame(ID,sales_df)
# Merge two data frame by "ID" and delete invalid information
type_sales_df<-merge(type_df,sales_df,by="ID")
type_sales_df<-na.omit(type_sales_df)

# Add frequency in type data frame
type_df<-count(type_df$unlist.type_list.)
# Delete invalid information for the last items 
type_df<-type_df[-nrow(type_df),]
# In descending order
type_df<-type_df[order(type_df$freq,decreasing=T),]

# Create a new data frame of sales in each type
sum_sales_list<-list()
for(i in 1:nrow(type_df)) {
  sum_sales<-0
  for(j in 1:nrow(type_sales_df)) {
    if(type_sales_df$unlist.type_list.[j]==type_df$x[i]) {
      sum_sales<-sum_sales+type_sales_df$unlist.sales_list.[j]
    }
  }
  sum_sales_list[i]<-sum_sales
}
sum_sales_df<-data.frame(type_df$x,unlist(sum_sales_list))
# Draw the pie chart of the number of type 
type_ePie<-ePie(type_df,size=c(650, 430))+eLegend(orient="vertical")+
  eTitle(title="各类型电影上榜数量饼图")
type_ePie
# Draw the pie chart of sales in different types
type_sales_ePie<-ePie(sum_sales_df,size=c(650, 430))+eLegend(orient="vertical")+
  eTitle(title="各类型电影票房饼图")
type_sales_ePie
  1. 统计电影主演的出现次数：
# ---------- Draw the cloud of actors ----------
# Extract the actor column and divide by /
actor<-movies$Actor
actor<-as.character(actor)
actor<-strsplit(actor,"/")

# Take the every actor of each movie and combine it into a new list
actor_list<-list()
for(i in 1:nrow(movies)) {
  add_list<-list(actor[[i]][1],actor[[i]][2],actor[[i]][3])
  actor_list<-c(actor_list,add_list)
}

# Build a new frame for actors 
actor_df<-data.frame(unlist(actor_list))
# Add frequency in actor data frame
actor_df<-count(actor_df$unlist.actor_list.)
# Delete invalid information for the head and tail
actor_df<-actor_df[-c(1:7),]
actor_df<-na.omit(actor_df)

# Extract the actor which appeared more than twice
actor_df<-actor_df[which(actor_df$freq >= 2),]
# In descending order
actor_df<-actor_df[order(actor_df$freq,decreasing=T),]
# Draw the cloud
wordcloud2(actor_df,size=0.3,minRotation=-pi/6,maxRotation=pi/6,rotateRatio=0.9)
  1. 机器学习
# Import packages 
library(ISLR)
library(e1071)
# Read the digital data
digital_movies<-read.csv("..\\data\\digital_movies.csv")
# Extract the data except column 1,3 and 10 as feature vector
x=digital_movies[,-c(1,3,10)]
# Extract the data from column 10 as outcome variable
y=digital_movies[,10]
# Build model: the kernel is polynomial kernel
model=svm(x,y,kernel="polynomial",gamma=if(is.vector(x)) 1 else 1/ncol(x),coef0=2)
summary(model)
## 
## Call:
## svm.default(x = x, y = y, kernel = "polynomial", gamma = if (is.vector(x)) 1 else 1/ncol(x), 
##     coef0 = 2)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  1 
##      degree:  3 
##       gamma:  0.1428571 
##      coef.0:  2 
## 
## Number of Support Vectors:  611
## 
##  ( 28 242 211 130 )
## 
## 
## Number of Classes:  4 
## 
## Levels: 
##  Bad Excellent Good Normal
# Test with train data
pred<-predict(model,x)
# Check accuracy
table(pred,y)
##            y
## pred        Bad Excellent Good Normal
##   Bad        86         0   14     32
##   Excellent   0        24    2      0
##   Good       29        14  428     84
##   Normal     28         0   34     97
# Calculate accuracy
table<-as.data.frame(table(pred,y))
count<-0
for(i in 1:nrow(table)) {
  if(table$pred[i] == table$y[i]) {
    count<-count+table$Freq[i]
  }
}
accuracy<-count/nrow(x)
accuracy
## [1] 0.728211
# Visualize (classes by color, SV by crosses):
plot(cmdscale(dist(x)),
     col=c("purple","red","orange","green")[as.integer(y)],
     pch=c("。","+")[1:200 %in% model$index+1])
legend(-0.96,-0.35,c("Bad","Excellent","Good","Normal"),
       col=c("purple","red","orange","green"),lty=1)

  1. 数据展示
# Interactive table for showing data 
DT::datatable(movies %>% head(100))

结论

本分析报告结论如下：

结论1

  • 由柱形图可以看出，评分与电影票房总体呈正相关趋势，评分越高的电影票房成绩越好；而评分与火爆程度（评论人数）总体也呈正相关趋势，评分越高的电影评论人数多，在社交媒体上热度较高。

结论2

  • 由饼图可以看出，动作、剧情、喜剧、冒险为当下比较受欢迎的电影类型，上榜数量多；而同样的，这些类型电影收获的票房也更高；但其中可以看出，喜剧在数量少于剧情但票房却高于剧情，说明喜剧类电影较剧情类而言更为卖座。

结论3

  • 由词云可以看出，古天乐、刘德华、甄子丹、黄渤、成龙上榜的次数最多，这与时下比较热门的“小鲜肉当道”的说法存在一定差异，老牌表演艺术家仍是电影主演的主力军。

结论4

  • 从支持向量机（SVM）模型的相关信息中可得知，共有611组支持向量（SV），在Bad、Excellent、Good、Normal四类中分别为28、242、211、130。由检验训练数据后的结果可以看出，在Bad、Excellent、Good、Normal四类中分别有86、24、428、97组数据分类正确，模型的准确率为72.82%。将结果可视化，可以看出数据总体分成四个簇，每个簇按照Bad、Normal、Good、Excellent从上至下分布，由于相互之间分布较为紧密，因此存在一定的分类误差。